University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 41


$$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$

Work Step by Step

We know that $$\frac{3\pi}{2}=\pi+\frac{\pi}{2}$$ Therefore, $$\sin\frac{3\pi}{2}=\sin\Big(\pi+\frac{\pi}{2}\Big)=\sin\pi\cos\frac{\pi}{2}+\cos\pi\sin\frac{\pi}{2}=0\times0+(-1)\times1=-1$$ $$\cos\frac{3\pi}{2}=\cos\Big(\pi+\frac{\pi}{2}\Big)=\cos\pi\cos\frac{\pi}{2}-\sin\pi\sin\frac{\pi}{2}=(-1)\times0+0\times1=0$$ Now return to the given quantity and apply the addition formula for sine: $$\sin\Big(\frac{3\pi}{2}-x\Big)=\sin\frac{3\pi}{2}\cos x-\cos\frac{3\pi}{2}\sin x$$ $$\sin\Big(\frac{3\pi}{2}-x\Big)=(-1)\times\cos x-0\times\sin x$$ $$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$
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